Two simple harmonic motion are represent by the following equations
`y_(1) = 10 sin (pi//4) (12 t + 1)`
`y_(2) = 5 (sin 3 theta t + sqrt3 cos 3 theta t)`
Here `t` is in seconds.
Find out the ratio of their amplitudes.What are the time period of the two motion?

Explanation

Given equation are `y_(1) = 10 sin (pi//4) (12t + 1)` (i)
`y_(2) = 5 (sin 3 pi t + sqrt3 cos 3 pi t)` (ii)
We recast these in the form of standard equation of SHM which is
`y = A sin (omega t + phi)` (iii)
Equation (i) may be written as
`y_(1) = 10 sin [(12 pi//4) + (pi//4)]`
`= 10 sin (3 pit + pi//4)]` (iv)
Comparing Eq (iv) with Eq (iii), we have
Amplitude of first SHM `= A_(1) = 10 cm//s and omega_(1) = 3 pi`
`:.` Time period of first motion `T_(1) = 2 pi // omega_(1) = 2pi//3pi = (2//3) s`
Let us put `5 = A_(2) cos phi` (v)
and `5sqrt3 = A_(2) sin phi` (vi)
Then `y_(2) = A_(2) cos phi sin (3 pi t + A_(2) sin phi cos 3 pi t)`
`= A_(2) sin (3 pi t+ phi)` (vii)
Squaring (5) and (6) and adding , we have
`A_(2) = sqrt([(5)^(2) + (5sqrt3)^(2)]) = 10 cm`
i.e., amplitude of second SHm is `10 cm` and time period of second AHM is `T_(2)`
`2 pi// omeag_(2) = 2 pi//3 pi = 2//3 s`
Thus, the ratio of amplitudes
`A_(1) : A_(2) = 1:1` and
perodic times are `T_(1) =T_(2) = (2//3) s`